Power Series
Last updated
Last updated
Try to think
Power series = Geometric Series
.
►Refer to Math24: Power series
Power series
is actually the Geometric series
in a more general and abstract form.
For easier to remember it, that could be simplified as:
In this function it's critical to know that: a_n
IS A CONSTANT NUMBER! NOT A VARIABLE !
Differentiate Power series
►Refer to Khan academy: Differentiating power series
We have 2 ways to differentiate series, they work same way:
One way is to expand the series with real numbers (1,2,3...) and take their derivatives:
(▲Note that this is constantly true for power series.)
Another way is to calculate the term's derivative and then plug in the real numbers (1,2,3):
Either way will do, it depends on the actual equation for you to choose which way you're gonna use.
Let's organize this function to make it clear:
Notice that: If we plug in the x=0
at beginning, everything will be 0
and we don't have anything to calculate.
So Let's keep the x
in the terms until the last step.
It's easier to expand the series with real numbers
, and we're to try 3 or 4 terms in this case.
Because at the end of it you'll notice, if we're doing Third Derivative
, then more than 3 terms will just bring more 0
s.
First we're to organize the function in the standard power series form:
And the constant number a_n
in the function is:
Let's plug in the real number (0,1,2,3,4) for n
to expand the series:
Based on that we can take the derivatives:
Now we've got the third derivative
, so let's plug in the x=0
and see what we get:
And that's the answer.
And now you know why in this case it's a waste to calculate more terms.
Integrate Power Series
Knowing that Integrate a series
can be turned to a series of Integrations of terms
:
Integrate the terms:
And we get a simple Geometric series
. So that we can apply the formula of calculating geometric series:
Let's apply the formula:
► Jump over to have practice at Khan academy.
Let's assume the function of the series above is f(x)
, and the series below is g(x)
It's clear to see that the g(x) is the
Antiderivativeof the
f(x)`.
So we just need to integrate the function of the f(x)
:
We see that the antiderivative
can represent the g(x)
, but only with the C
in it:
So we need to solve for C
. The easiest way is to plug in 0
for g(x)
:
Then the answer is:
Set the two series as f(x)
and g(x)
.
It's clear that g(x) = -f'(x)
.
So by differentiate f(x)
we will get:
Then -f'(x)
would be:
Solve:
Solve:
Solve:
Solve:
Solve: